General statistics question: low probability, high variance

[NotEnoughHeat]

Hi, I'm back with another beginner level question that my pride makes it a little embarrassing to ask. But it's much better to ask it now, then wait till I'm too embarrassed to ask at all!

This is something I've I haven't heard addressed explicitly in the way I'm going to put it since it's pretty simple. I'm pretty sure I already know the answer, but would just like someone to confirm and add comments (however, I'd would find it more interesting if I didn't have it right).

If you have two different bets (in general, not blackjack bets in particular) with the same edge, does the bet with the lower probability always have a higher variance?

I've always took the variance/standard deviation of a game to refer to the units won or loss, but I suppose authors sometimes refer to other related variables or qualify them in different ways (win & loss per units bet or per hand). So my thinking is that it's much easier for a 1 in 10 bet to be off of expectation than a 50/50 one. For example imagine both have 0 edge and bet is wagered 10 times. It seems to me that it's more likely that you'll be down 4 units in the 1 in 10 than being down 4 units in the 50/50 proposition. I could go on to describe my thinking further, but the rest is probably either redundant or obvious.

Please let me know if I'm missing something, thanks!

[Three]

So my thinking is that it's much easier for a 1 in 10 bet to be off of expectation than a 50/50 one. For example imagine both have 0 edge and bet is wagered 10 times. It seems to me that it's more likely that you'll be down 4 units in the 1 in 10 than being down 4 units in the 50/50 proposition. I could go on to describe my thinking further, but the rest is probably either redundant or obvious.

After 10 bets with a 10 to 1 payout you either are:
-10 bets
+1 bet
+12 bets
+23 bets
+34 bets
+45 bets
+56 bets
+67 bets
+78 bets
+89 bets
+100 bets

Most of the time you will be -10 bets to +12 bets with by far most common outcomes for 10 bets as -10 bets and +1 bet. After 0 bets you can't be down 4 units exactly. Of course with no edge you know the average will be 0 units but that isn't one of the possibilities either. It does give you the sense of where the main frequencies fall.

For the even money bet:
-10 bets
-8 bets
-6 bets
-4 bets
-2 bets
EVEN
+2 bets
+4 bets
+6 bets
+8 bets
+10 bets

The frequency distribution will peak at the center but have a much wider spread of frequent outcomes. Both sides of the 0 bet average on this frequency would be mirror images. If you notice the range of results in the even money situation fit within the 3 most common outcomes of the 11 possible outcomes for the 10 to 1 bet.

If you have two different bets (in general, not blackjack bets in particular) with the same edge, does the bet with the lower probability always have a higher variance?

This is a generalization that doesn't always hold true unless both bets have the same house edge (as is specified in your question) it should always be true. I hope some of the geniuses on the site will correct me if I am wrong. It gets trickier when you look at bets that have a pay table rather than a flat rate standard payout. Like a slot machine with a big progressive. For the odds to be the same as another slot machine without the progressive the non-progressive will pay better on at least some of the payoffs you are likely to hit or hit the lower payouts more frequently to make up for the progressive payout you will probably never hit. The more frequent you get better payoffs on the more likely wins or the increase in likelihood of the smaller wins to get the same HE the lower the variance should be. It is getting pretty complicated to generalize at that point but I hope you get the idea.

[SiMi]

Hi, NEH!

If I follow your question, I think you've got it right. I'm assuming your scenario is that one bet has a 10% probability of winning (on average, it will win 1 out of 10 times) and the other has a 50% probability of winning (on average, it will win 1/2 of the time).

I'm also assuming that the bet on the 10% probability game will pay more than the 50% game if it wins. We could assume, for example, that the 50% game will pay 1 unit for every unit you bet if you win and the 10% game will pay 9 units for every unit you bet if you win. If we assume that, then, ON AVERAGE, every 10 rounds of the game would look like this if you bet 1 unit each time:

50% game pays 1:1
Trial: 1 2 3 4 5 6 7 8 9 10 Net Win
Result: +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 0 units (win every other time like flipping a coin)

10% game pays 9:1
Trial: 1 2 3 4 5 6 7 8 9 10 Net Win
Result: -1 -1 -1 -1 -1 -1 -1 -1 -1 +9 0 units (lose 9 times and then win a big one-kinda like Video Poker!)

In both cases, your edge (or 'average' or 'EV' or 'Mu' or 'mean') is the same (0%). However, when we calculate the Variance (V), we get different answers because:

V = Epsilon [ (outcomes - average)^2 ] / No. of Trials (Epsilon means "take the sum of" and the caret symbol ( ^ ) means "raise to the power of")

In English: Variance equals the sum of: The square of: The actual outcomes minus the average outcome all divided by the number of trials. So, we would take EACH OUTCOME and subtract the mean value from it (0 in both cases) and then square it. Then, we would add up each value and then divide that total by the number of trials. Here, we ran 10 trials.

So, for example, with the 50% game, the first outcome was for us to win 1 unit while the average outcome is to win 0 units. This would be our first term in calculating V and we could show it as: (+1 - 0)^2 = (1)^2 = 1. In English: The square of: 1 minus zero is 1 squared which is simply 1. We need to do that to EACH OUTCOME and sum up all these values and then divide the total by the number of trials or, in this case, 10.

With the 50% game, we would therefore have:

V = [(+1 - 0)^2 + (-1 - 0)^2 + (+1 - 0)^2 + (-1 - 0)^2 + (+1 - 0)^2 + (-1 - 0)^2 + (+1 - 0)^2 + (-1 - 0)^2 + (+1 - 0)^2 + (-1 - 0)^2] / 10 = 10/10 = 1

With the 10% game, we would have:

V = [(-1 - 0)^2 + (-1 - 0)^2 + (-1 - 0)^2 + (-1 - 0)^2 + (-1 - 0)^2 + (-1 - 0)^2 + (-1 - 0)^2 + (-1 - 0)^2 + (-1 - 0)^2 + (+9 - 0)^2] / 10 = (9 + 81) / 10 = 90/10 = 9

Once we have the Variance, the Standard Deviation (SD) is simply the square root of the Variance. So:

50% game: V = 1 SD = sqrt(1) = 1
10% game: V = 9 SD = sqrt(9) = 3

So, you can see that in this example (which I set up in order to show the Long Run values), the 10% game has 9 times the Variance (which simply reflects the Range) and 3 times the SD of the 50% game (which is a REALLY boring game).

I was aided in answering this question by my good friend, Skull21, who is a humble genius!

I hope this helps and I didn't screw it up too bad. If so, it's all my bad and not Skull's. If you need more, please let me know and I'll try again.

Best,
SiMi

[NotEnoughHeat]

Thank you both for your thoughtful and illustrative responses. Even though I already had the right answer, I feel like I understand it that much clearer now. I also hope that this thread will benefit others as well.

Also, just because I like to nit pick (hope it isn't too annoying) I'm going to point out one thing in T3s post. The particular example I was asking about intended a 1 in 10 bet to refer to a bet with a 10% probability of winning (and 90% for losing). The first game, as described in the post, with a 10 to 1 payout is actually a positive expectation game (10%) and the average won (cumulative EV) would be 1 unit. I don't know whether or not T3 intended to imply that the game he described has no edge, but it's besides the point and everything that was said, as far as I can tell, is 100% correct.

[NotEnoughHeat]

Reading Simi's and T3s prior posts, I'd say one could pretty much take their words as gospel for the most part. But I want to finish the book I'm reading "Gone with the Wind" in japanese before I put it the time to absorb what they just said. Is there a relavance to blackjack? Or is this another one of those RofR deals?

Well it's a statistics/probability question, so just by virtue of that I think it has some relevance. Whether or not it's relevant enough for the forum might be a different question though. I think the question definitely has value for the blackjack player, as does a lot of general statistics/probability stuff, though indirect.

For me personally however, this question relates to my blackjack play when I deal with side bets. The side bet I count contributes a significant proportion of my EV from blackjack overall. The probability of hitting any payout is about 11-12%. If I am right to think that it's high variance then I'm adding a decent amount of volatility to my game by playing it and will have to weigh that in my decision in how much action I want to give it. However, if it's variance was the exact same as normal blackjack counting then I would be crazy not to max it out, considering the edge is much higher.

I'm not really sure what you mean by 'one of those RofR deals'. I thought the threads about risk of ruin, kelly and bankroll dealt with subjects with a lot of attention and importance to the AP community and literature.

[SiMi]

Hey, moses!

I got a kick out of your comments. There are definitely 2 types of people who play AP BJ. But, it seems to me, without Ed Thorp and his kind, we could remove the "AP" from "AP BJ."

Take care,
SiMi

[NotEnoughHeat]

Hey Simi. I too enjoy reading your very informative posts. Once I understood the purpose of Notenoughheats question. It makes perfect sense. Somehow I was under the impression he was new to the game. I think you'd agree few new players are advanced enough to be concerned with side counts. So my apologies to NEH.

No worries. lol I take it as a compliment that you think I'm not new to the game so I hate to burst the bubble. It really depends what you mean by 'new' and 'the game' when considering when I fit the description of being new to the game. I've been playing blackjack since I was old enough to frequent a casino (in my 3rd year) and I've only fourth or fifth month of counting. Also, I'm not actually using a side count. I'm not sure whether you misread 'side bet' as 'side count' or assumed that I was using a side count to attack my side bet. The count I use for normal blackjack has a good enough correlation to the side bet that I can use it to determine when it's profitable to bet.

Although I might considered new to the game, I think I am pretty well read on the game for noobie. Hopefully some of 'nice to know' questions find some good practical value.

[SiMi]

Hi, Moses,

God, if only I could go back to the OLD DAYS! Single deck games, deeply dealt were common and you COULD look pretty smart. But, the truth is that I would NEVER have figured out what Thorp came up with on my own. Not in a million years. Wonging in and out was also a major advance for me thanks to "Stanford." Don's I18 is HUGE to counters and there are so many other smart guys who have pushed us ahead over the years.

You definitely do NOT need to know what Don S knows to be a successful counter just as you don't need to know how to rebuild an engine to drive a car well. Some of us just like to know what makes things tick and so we go after the nerdy aspects of the game because it's fun to us. Every now and then a Norm or a Don will come up with a great idea and we all benefit when they share it with us. I play several times a week and I use everything I can to keep an edge. It's a blend of theory and practice for me and I love it that way.

As you say, to each his own. When I see a question such as NEH asked, I assume the author is sincerely trying to sort something out and needs some help. If I can help, I try my best because it's what I would want if I were in their shoes.

You never know... One day, one of these guys might come up with something that will help all of us. I'm hoping they'll be willing to share when they do, in part, because someone helped THEM when they needed it.

Best of luck to you!
SiMi

[SiMi]

Moses

Yep, all I had was "The Book" from the library and I was bounced out a lot, too. I played with another guy a lot and, thankfully, they had to bounce BOTH of us at a time. That tended to make me feel a little safer on the way to the door. I was really good at tracking Aces and he was a smooth talker and was always hitting on the ladies. We had some fun!

I remember playing with actual silver dollars where the min bet was $1 and max was $100. I didn't have the balls/BR to spread more than 10:1 at the time unless I was a little drunk. I guess shoe games were around in 1979 but I don't remember seeing them in Vegas or Reno. I only played SD and made some decent money as far as I was concerned.

Like you, I had no other resources and really didn't know my a$$ from a hole in the ground about the game when I started playing with real money. Personal Computers didn't exist yet so we were just trusting Thorp. It was fun and after you got the boot, you could just walk a few feet and start again. Free drinks, too!

It feels a lot different today and my style of play is a LOT different. I'm starting to feel pretty old. The only good news is that I'm so old, I don't fit the mold of the 'typical' counter!

Have a great evening, Moses!
SiMi

[Nyne]

SiMi shows you the math for standard deviation and variance, which (at a glance) looks right. I'll add that, for a simple 2 outcome (win/lose) bet, the variance is approximately equal to the payoff odds. For a 0% edge bet, it's exactly equal to the payoff odds, as SiMi showed (the 1:1 bet has a variance of 1, the 9:1 bet has variance of 9.) As the edge moves away from zero, the estimate is less accurate, but it's good for a quick mental estimate when you can't break out your calculator for whatever reason.